Swagger

Question

我有一个 swagger 定义，我正在使用 XML 评论来描述请求和响应并提供有意义的示例。示例模型如下：

/// <summary>Created item information.</summary>
public class ItemCreated
{
    /// <summary>Outcome of the item being created.</summary>
    /// <value>The outcome.</value>
    /// <example>Item has been Created</example>
    public string Outcome { get; set; }

    /// <summary>Unique item reference.</summary>
    /// <value>The Item reference.</value>
    /// <example>6001002982178</example>
    public string Reference { get; set; }

    /// <summary>The external reference for the package.</summary>
    /// <value>The carrier reference.</value>
    /// <example>5558702516</example>
    public string ExternalReference { get; set; }

    /// <summary>The items documents.</summary>
    /// <value>The items documentation.</value>
    /// <example>???</example>
    public List<Documentation> Documents { get; set; }
}


/// <summary>Item documentation information.</summary>
[JsonObject("Documentation")]
public class Documentation
{
    /// <summary>The document in base64 format.</summary>
    /// <value>The document base64 string.</value>
    /// <example>JVBERi0xLjMNCjEgMCBvYmoNCjw8DQovVHlwM...</example>
    [JsonProperty("Document")]
    public string Document { get; set; }

    /// <summary>The type of the document.</summary>
    /// <value>The documentation type.</value>
    /// <example>ITEMDOCUMENT_SLIP1</example>
    public DocumentationType Type { get; set; }

    /// <summary>Document format.</summary>
    /// <value>The format of the document.</value>
    /// <example>Pdf</example>
    public string Format { get; set; }

    /// <summary>The document resource uri.</summary>
    /// <value>The link to the document resource.</value>
    public string Link { get; set; }
}

大摇大摆显示为：

我需要的是展示一个更复杂的响应示例。您可以在此处看到其中一个属性是一个数组——我想显示多种文档类型——所以有多个数组项。我如何显示这个？ "" 节点不允许在一个数组中包含多个示例。

感谢您提前提出任何意见！

Answer 1

一种方法是使用架构过滤器：

[SwaggerSchemaFilter(typeof(ItemCreatedSchemaFilter))
public class ItemCreated
{
    ⋮
}

public class ItemCreatedSchemaFilter : ISchemaFilter
{
    public void Apply(OpenApiSchema schema, SchemaFilterContext context)
    {
      if (schema.Type == typeof(ItemCreated))
      {
        schema.Example = new OpenApiObject
        {
            // Other Properties
            ["Documents"] = new OpenApiArray
            {
                new OpenApiObject
                {
                    // Other Properties
                    ["Document"] = new OpenApiString("JVBERi0xLjMNCjEgMCBvYmoNCjw8DQovVHlwM..."),
                    ["Format"] = new OpenApiString("Pdf")
                },
                new OpenApiObject
                {
                    // Other Properties
                    ["Document"] = new OpenApiString("Something else"),
                    ["Format"] = new OpenApiString("Something else")
                }
            }
        };
      }
    }
}

参考：Apply Schema Filters to Specific Types

记得将新过滤器添加到 SwaggerGen 选项中：

services.AddSwaggerGen(c =>
{
     c.SchemaFilter<ItemCreatedFilter>();

     ...
}

Swagger - 如何显示更复杂的响应示例 - ASP.net Core Web API

Swagger - how to show a more complex response example - ASP.net Core Web API

c#

asp.net-web-api

asp.net-core

openapi